Complex Analysis/Appendix/Proofs/Triangle Inequality

Let $z$ and $w$ be complex numbers. Since we have:

$|z+w{|}^2$	$=(z+w)\overline{(z+w)}=(z+w)(\overline{z}+\overline{w})$
	$=|z{|}^2+z\overline{w}+\overline{z\overline{w}}+|w{|}^2$
	$=|z{|}^2+2\text{Re }(z\overline{w})+|w{|}^2$
	$\le |z{|}^2+2|z||w|+|w{|}^2$
	$=(|z|+|w|{)}^2$

the triangular inequality follows after taking the square root of both sides. Note here we used the properties:

$\text{ Re}(z)\le |z|$, $|z|=|\overline{z}|$ and $z+\overline{z}=2\text{Re }(z)$.

Also, the induction shows:

$\left|\sum _1^n{z}_k\right|\le \sum _1^n|z_k|$